Question

two resistors of 6 ohm and 3 ohm are connected in parallel . four such combination are connected in series. calculate the equivalent resistance thus formed.

Answer 1

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Since 6 ohm and 3 ohm are in parallel hence there resistance is R1.1R1=13+16=2+16=12R1=2 ohmSince 4 such combinatins arein series thereforenet equivalnet resistance is R.R=2+2+2+2=8 ohm

Answer 2

Answer:

The equivalent resistance of parallel combination connected in series is 8Ω.

Explanation:      

Given two resistors of 6 ohm and 3 ohm.

Let $R_1=6 \Omega ~\&~R_2=3 \Omega$

When two resistors are connected in parallel, then the equivalent resistance of the combination is given by

$\dfrac{1}{R} =\dfrac{1}{R_1}+\dfrac{1}{R_2}$

Substituting the given values,

$\dfrac{1}{R} =\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{1+2}{6}=\dfrac{3}{6}$

$\implies R=\dfrac{6}{3}=2\Omega$

Given four such combinations are connected in series.    

When the resistors are connected in series, then the equivalent resistance of the combination is given by sum of the individual resistances of resistors.

Therefore, the equivalent resistance of the combination is

$R_{eq} = 2+2+2+2~ \mbox{or}~ 2\times 4=8\Omega$

Therefore, the equivalent resistance of parallel combination connected in series is 8Ω.